What are the solutions to the equation d_{1}-3x^2+4x=0 ?

The solutions to the equation d_{1}-3x^2+4x=0 are x=-(-2+sqrt(3d_{1)+4})/3 ,x=(sqrt(3d_{1)+4}+2)/3

Find the zeros of d_{1}-3x^2+4x=0

The zeros of d_{1}-3x^2+4x=0 are x=-(-2+sqrt(3d_{1)+4})/3 ,x=(sqrt(3d_{1)+4}+2)/3

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d_{1} - 3 x^{2} + 4 x = 0

x = - \frac{- 2 + 3 d _{1} + 4}{3}, x = \frac{3 d _{1} + 4 + 2}{3}

Solution steps

d_{1} - 3 x^{2} + 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- 3 x^{2} + 4 x + d_{1} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 3 ) d _{1}}{2 ( - 3 )}

Simplify

4^{2} - 4 (- 3) d_{1} : 2 4 + 3 d_{1}

x_{1, 2} = \frac{- 4 \pm 2 4 + 3 d _{1}}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 4 + 2 4 + 3 d _{1}}{2 ( - 3 )}, x_{2} = \frac{- 4 - 2 4 + 3 d _{1}}{2 ( - 3 )}

x = \frac{- 4 + 2 4 + 3 d _{1}}{2 ( - 3 )} : - \frac{- 2 + 3 d _{1} + 4}{3}

x = \frac{- 4 - 2 4 + 3 d _{1}}{2 ( - 3 )} : \frac{3 d _{1} + 4 + 2}{3}

The solutions to the quadratic equation are:

x = - \frac{- 2 + 3 d _{1} + 4}{3}, x = \frac{3 d _{1} + 4 + 2}{3}

x, p = b - x^{2} a

(2 m - 5) \cdot x^{2} + x + 4 = 0

4.9 x^{2} = 78.4

x, x^{2} - y = y^{2} - x

21 x^{2} + 21 x - 254 = 0

What are the solutions to the equation d_{1}-3x^2+4x=0 ?
The solutions to the equation d_{1}-3x^2+4x=0 are x=-(-2+sqrt(3d_{1)+4})/3 ,x=(sqrt(3d_{1)+4}+2)/3
Find the zeros of d_{1}-3x^2+4x=0
The zeros of d_{1}-3x^2+4x=0 are x=-(-2+sqrt(3d_{1)+4})/3 ,x=(sqrt(3d_{1)+4}+2)/3